Understanding arithmetic sets - GMAT Quantitative

comprises the set of elements common to all three sets. However, since is the set of all even integers and comprises the set of all odd integers, no element can be common to all three sets. The correct response is 0.

comprises the set of all odd integers except 1; comprises the set of all integers of the form , a natural number. Therefore, any number that is not in the union of these two sets must be in neither one.

, so is even or 1 (although 1 is not a choice). We can eliminate odd choices 147, 149, and 151 immediately.

, so we determine which number cannot be expressed as , a natural number.

148 is elminated, since it is two less than a multiple of 3. 150 is the correct choice.

Add each element of to each element of .

= {1 + 3, 1 + 4, 1 + 6, 4 + 3, 4 + 4, 4 + 6} = {4, 5, 7, 8, 10}

is the set of all elements that are in both and . So in this case the elements that are the same for both sets are , Note that the order that you present the elements in your set doesn't matter. What matters is that you don't exclude any necessary elements, or add any that don't belong.

There are three choices for (1, 2, and 3), and similarly there are three choices for (also 1, 2, and 3). Together there are possible functions from to . Remember to multiply, NOT add.

are the sets of all positive integers that are one greater than a multiple of five, four, and three, respectively. Therefore, for a number to be in all three sets, and subsequently, , the number has to be one greater than a number that is a multiple of five, four, and three. Since , the number has to be one greater than a multiple of 60. The first four numbers that fit this description are 1, 61, 121, and 181, the last of which is the correct choice.

comprises the set of elements common to all three sets. Only 5 fulfills that condition, so the correct choice is 1.

The numbers in Region I are exactly the natural numbers . All natural numbers are integers, and the integers are closed under subtraction, so cannot fall in Region IV or Region V. Also, since it is given in the problem that , it follows that , so the difference cannot be in Region II (the only whole number that is not a natural number is ).

It is possible for to be in Region I. Example:

It is possible for to be in Region III (the integers that are not whole numbers, or the negative integers). Example:

can fall in either of two different regions.

The subset relation is transitive, so and together imply that .

Since all three of are subsets of , then any element of any of those four sets is an element of . Contrapositively, any nonelement of cannot be an element of any of .

However, it is possible for a nonelement of to be an element of if . A simple example:

and .

, , and .

Order and repetition do NOT change a set. Therefore, the set we want to describe contains the numbers 1, 3, and 4. The only set that doesn't contain all 3 of these numbers is , so it is the set that does not equal the rest of the sets.

We need to add 3 to every element in .

Then:

If heads comes up on the coin, the number on the die is recorded. This can be any element of the set .

If tails comes up on the coin, half the number on the die is recorded, and any fraction is thrown out. Half the elements of the set comprise the set , but since we are throwing out the fractional parts, this becomes the set .

The sample space is the union of these two sets, which is .

If heads comes up on the coin, the number on the die is recorded. This can be any element of the set .

If tails comes up on the coin, twice the number on the die is recorded. This can be twice any element of the set - that is, any element of the set .

The sample space is the union of these two sets:

The intersection of and is the set of all elements in that are also in - namely, . The union of this set and is the set of all elements in one or the other set, or .

The set has seven elements.

You only need to look for the portion of the line with the greatest negative slope, which is that which represents Friday to Saturday.

728 is divisible by 7, but not 3 or 5; therefore, 728 is in but not or . To be in the same region, a number must also be in but not or - that is, divisible by 7 but not 3 or 5.

510 and 595 can be eliminated as multiples of 5 (from the last digit); 777 can be eliminated as a multiple of 3 (digit sum is 21).

Now let's look at the two reminaing choices.

736 is not divisible by 7, since .

476 is not divisible by 3 or 5, but it is divisible by 7:

476 is the correct choice.

The number of subsets that can be formed from a set with elements is . However, and , so there is no integer for which . Therefore, a set with exactly 768 elements cannot exist.

1,225 is divisible by 5 (last digit is 5). It is a perfect square, since . It is not a perfect cube, however, since .

Therefore, 1,225 is an element in and , but not . We are looking for an element in and , but not - that is, a multiple of 5 and a perfect square but not a perfect cube.

By looking at the last digits, we can immediately eliminate 1,764 and 4,356, since neither is a multiple of 5. We can eliminate 15,625, since it s a perfect cube - .

Of the two remaining numbers, 3,375 is not a perfect square, since

3,025 is a perfect square:

3,025 is not a perfect cube:

3,025 is a multiple of 5, as can be seen from the last digit.

This is the correct choice.

So the intersection of A and B is the numbers that are in both A and B.

Thus